Block stacking problem

Block stacking problemGilles CazelaisSuppose we have a stack of n identical blocks of unit length and we try to balance them on the edgeof a table. What is the largest possible offset distance from the table’s edge?Let d n be the maximum offset distance of a stack of n blocks.Tabled nWhen the largest offset is obtained, the center of mass of the n blocks must lie right above the table’sedge and the center of mass of the n − 1 top blocks must lie right above the edge of the bottom block.n blocksn − 1 blocksd nd n−1TableTabled n−1 + 1 2Let M be the mass of a single block. By computing the total moment of the n blocks with respect tothe right edge we obtainnMd n = M(d n−1 + 1 2 ) + (n − 1)Md n−1.We can solve for d n to obtain the recurrence relationIt is easy to see that d 1 = 1 2 .d n = d n−1 + 12n .Tabled 1 = 1 2By iteration we obtaind 2 = d 1 + 12 · 2 = 1 2 + 1 4 = 1 (1 + 1 )2 2d 3 = d 2 + 12 · 3 = 1 2 + 1 4 + 1 6 = 1 2d 4 = d 3 + 12 · 4 = 1 2 + 1 4 + 1 6 + 1 8 = 1 2(1 + 1 2 + 1 )3(1 + 1 2 + 1 3 + 1 ).4

We see that in general we have the following formula.d n = 1 2(1 + 1 2 + 1 3 + · · · + 1 )nThis result could be proved by mathematical induction.Since we know that the harmonic series1 + 1 2 + 1 3 + 1 4 + · · ·diverges to infinity, we get the surprising fact that the offset distance satisfieslim d n = ∞.n→∞This means that d n can become arbitrarily large provided that we choose n large enough.⋆ ⋆ ⋆How many blocks do we need so that the left edge of the top block extends beyond the edge of thetable? Sinced 3 = 1 (1 + 1 2 2 + 1 )= 11 and d 4 = 1 (1 + 1 3 122 2 + 1 3 + 1 )= 254 24we see that at least 4 blocks are needed.Tabled 3 < 1Tabled 4 > 1It is interesting to observe that sinced 52 = 1 2(1 + 1 2 + 1 3 + · · · + 1 )≈ 2.27,52if we use a deck of 52 cards instead of blocks, we can obtain an offset distance of about 2.27 times thelength of a card.Tabled 52 ≈ 2.27Gilles Cazelais. Typeset with LATEX on July 14, 2006.